Earth Science Asked by J-P on September 5, 2021
How should I interpret the hourly precipitation forecasts?
Most of the time, when there is rain in the forecast, the maximum Probability of Precipitation (PoP) in the hourly forecast is about the same as the daily PoP.
My problem is that the hourly PoP forecast usually has several hours at that maximum probability. If I naively add the odds of getting wet for each hour, I always get a total probability that is much higher than the daily PoP value.
For example, take an hypothetical day where the daily forecast has a 50% PoP. For the same day, the hourly forecast has 4 hours with a 50% PoP for each hour. If I compute the chance that I get rain in at least one of the 4 hours, I get 93% PoP (15 out of 16).
How can I reconcile the total daily PoP calculated from the hourly forecast with the actual daily PoP?
The hours are not independent of each other. For example, suppose there's a chance of a hurricane hitting the area. If it hits, it will be rainy every hour of the day. If it misses, it will not rain at all. On the other hand, thunderstorms are more independent: getting one at 2pm doesn't increase the chance of getting another one at 4pm.
To the extent that the hours are independent you can calculate the daily probability as you described; however, it's more likely that they are dependent on each other, especially in winter.
Source: https://www.insidescience.org/news/how-do-daily-weather-forecasts-relate-hourly-forecasts-it-depends
Answered by Richard H Downey on September 5, 2021
The probability of precipitation is most likely to mean the proportion of models in an ensemble or weather models in which precipitation was observed at a particular location over a particular time period.
If you want to work out the probability of it raining during the day, the best approach would be to work out one minus the probability that there was no rain in hour one and there was no rain in hour two and there was no rain in hour three... Mathematically if $x_i$ represents the presence or absence of rain in hour $i$, then
$P(rain) = 1 - prod_{i=1}^{24} (1 - P(x_i = 1))$
However, this assumes that the probabilities of rain in each hour are independent, which is unlikely to be true. To deal with this, we would need to introduce conditional probabilities (which are not supplied by the forecast), i.e.
$P(rain) = 1 - (1 - p(x_1=1)times(1 - p(x_2=1|x_1))times(1 - p(x_3=1|x_1,x_2))times cdot times(1 - p(x_{24}=1|x_1,x_2,ldots,x_{22},x_{23})$
So unfortunately you will not be easily able to reconcile the daily and hourly probabilities as this would require the conditional probabilities (or at least a model of the dependence), which probably is not supplied in the forecast.
In the particular example given in the question, if there were four hours that had a probability of rainfall of 50% then this could mean that 50 out of 100 models in the forecast ensemble had rain that lasted all four hours and the other 50 were dry. However, it could have meant that 50 models have rain for the first two hours and dry after that, and the other fifty were dry for two hours and then had rain for two hours. In the first case, the probability of there being rain at some point during the four hours is 50% (as half of the models say there will be no rain at any point) whereas in the second case, the chance of rainfall at some point will be 100% as all the models agree that it will rain during the four hour period, they just disagree as to precisely when it will happen. The daily forecast works this out by looking at the proportion of models that actually gave some precipitation at some point during the day, but the hourly probabilities in isolation don't give enough information to work out what happened throughout the day in individual model runs.
Answered by Dikran Marsupial on September 5, 2021
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